How to solve this equation algebraically Solve the following simultaneous equations on the set of real numbers: 
\begin{cases}x^2 + y^3 = x+1 \\ x^3+y^2=y+1\end{cases}
Thanks for helping!
 A: Subtract the two equations to get 
$(x^2 -y^2) -(x^3 -y^3)-(x-y)=0$
$\Rightarrow (x-y)(x+y-x^2-y^2 -xy-1)=0$
So, either $x-y=0$ or $x+y-x^2-y^2 -xy-1=0$


*

*For the first case, 
We have $x^3+x^2-x-1=(x-1)(x^2 -1)=(x-1)^2 (x+1)=0 \Rightarrow x=1 $ or $x=-1$
So, $x=y=1$ or $x=y=-1$

*And for the second case,
$x+y-x^2-y^2 -xy-1=-\frac{1}{2} \cdot [(x-1)^2 +(y-1)^2 +(x+y)^2]=0$ 
$\Rightarrow x=1, y=1 $ and $x=-y$, which is not possible at the same time. 
So, this case yields no solutions.
Thus the only solutions are $x=y=1$ and $x=y=-1$
A: Hint:
Eliminate one of the unknowns in a way to get a polynomial.
From the second equation,
$$y^2-y+1=2-x^3$$ and multiplying by $y+1$ and using the first equation,
$$(2-x^3)(y+1)=(y^2-y+1)(y+1)=y^3+1=2+x-x^2,$$
and
$$y=\frac{2+x-x^2}{2-x^3}-1=\frac{x-x^2+x^3}{2-x^3}.$$
Then
$$\left(x-x^2+x^3\right)^3=(1+x-x^2)(2-x^3)^3.$$
A: The corresponding graphs of the two curves are symmetrical with respect to the diagonal $y=x$ which is clear via the the transformation $(x,y)\to(y,x)$.
It follows the common points are such that $x=y$ so we have the equation to solve $$x^3+x^2-x-1=(x-1)(x+1)^2=0$$ which have the solutions $1$ and $-1$ with $-1$ as a double root.
Thus the solutions are $(1,1)$ and $(-1,-1)$.
